Run-time Environment and Program Organization

1. Run-time Environment and Program Organization

2. Outline • Management of run-time resources • Correspondence between static (compile-time) and dynamic (run-time) structures

3. Run-time Resources • Execution of a program is initially under the control of the operating system • When a program is invoked: • The OS allocates space for the program • The code is loaded into part of the space • The OS jumps to the entry point (i.e., “main”)

4. Low Address Code Memory High Address Memory Layout Other Space

5. What is Other Space? • Holds all data for the program • Other Space = Data Space • Compiler is responsible for: • Generating code • Orchestrating use of the data area

6. Assumptions about Execution • Execution is sequential; control moves from one point in a program to another in a well-defined order • When a procedure is called, control eventually returns to the point immediately after the call Do these assumptions always hold?

7. Activations • An invocation of procedure P is an activation of P • The lifetime of an activation of P is • All the steps to execute P • Including all the steps in procedures P calls

8. Activation Trees • Assumption (2) requires that when P calls Q, then Q returns before P does • Lifetimes of procedure activations are properly nested • Activation lifetimes can be depicted as a tree

9. g f g Example class Main { int g() { return 1; } int f() {return g(); } void main() { g(); f(); } } Main

10. Example 2 class Main { int g() { return 1; } int f(int x) { if (x == 0) { return g(); } else { return f(x - 1); } } void main() { f(3); } } What is the activation tree for this example?

11. Notes • The activation tree depends on run-time behavior • The activation tree may be different for every program input • Since activations are properly nested, a stack can track currently active procedures

12. Example class Main { int g() { return 1; } int f() { return g(); } void main() { g(); f(); } } Main Stack Main

13. g Example class Main { int g() { return 1; } int f() { return g(); } void main() { g(); f(); } } Main Stack Main g

14. g f Example class Main { int g() { return 1; } int f() { return g(); } void main() { g(); f(); } } Main Stack Main f

15. g f g Example class Main { int g() { return 1; } int f() { return g(); } void main() { g(); f(); } } Main Stack Main f g

16. Low Address Code Memory Stack High Address Revised Memory Layout

17. Activation Records • The information needed to manage one procedure activation is called an activation record(AR) or frame • If procedure F calls G, then G’s activation record contains a mix of info about F and G.

18. What is in G’s AR when F calls G? • F is “suspended” until G completes, at which point F resumes. G’s AR contains information needed to resume execution of F. G’s AR also contain: • Actual parameters • Space for G’s return value • Pointer to the previous activation record • The control link; points to AR of caller of G • Other temporary values • Local variables

19. Example 2, Revisited class Main { int g() { return 1; } int f(int x) { if (x == 0) { return g(); } else { return f(x - 1); (**)} } void main() { f(3); (*)} } AR for f:

20. Main Stack After Two Calls to f f (result) 3 (*) f (result) 2 (**)

21. Notes • Main has no argument or local variables and its result is never used; its AR is uninteresting • (*) and (**) are return addresses of the invocations of f • The return address is where execution resumes after a procedure call finishes • This is only one of many possible AR designs • Would also work for C, Pascal, FORTRAN, etc.

22. The Main Point The compiler must determine, at compile-time, the layout of activation records and generate code that correctly accesses locations in the activation record

23. Backup Slides

24. Main f (result) 3 (*) f 1 2 (**) Example The picture shows the state after the call to 2nd invocation of f returns

25. Discussion • The advantage of placing the return value 1st in a frame is that the caller can find it at a fixed offset from its own frame • There is nothing magic about this organization • Can rearrange order of frame elements • Can divide caller/callee responsibilities differently • An organization is better if it improves execution speed or simplifies code generation

26. Discussion (Cont.) • Real compilers hold as much of the frame as possible in registers • Especially the method result and arguments

27. Globals • All references to a global variable point to the same object • Can’t store a global in an activation record • Globals are assigned a fixed address once • Variables with fixed address are “statically allocated” • Depending on the language, there may be other statically allocated values

28. Low Address Code Memory Static Data Stack High Address Memory Layout with Static Data

29. Heap Storage • A value that outlives the procedure that creates it cannot be kept in the AR • Bar foo() { return new Bar } The Bar value must survive deallocation of foo’s AR • Languages with dynamically allocated data use a heap to store dynamic data

30. Notes • The code area contains object code • For most languages, fixed size and read only • The static area contains data (not code) with fixed addresses (e.g., global data) • Fixed size, may be readable or writable • The stack contains an AR for each currently active procedure • Each AR usually fixed size, contains locals • Heap contains all other data • In C, heap is managed by malloc and free

31. Notes (Cont.) • Both the heap and the stack grow • Must take care that they don’t grow into each other • Solution: start heap and stack at opposite ends of memory and let the grow towards each other

32. Low Address Code Memory Static Data Stack High Address Heap Memory Layout with Heap

33. Data Layout • Low-level details of machine architecture are important in laying out data for correct code and maximum performance • Chief among these concerns is alignment

34. Alignment • Most modern machines are (still) 32 bit • 8 bits in a byte • 4 bytes in a word • Machines are either byte or word addressable • Data is word aligned if it begins at a word boundary • Most machines have some alignment restrictions • Or performance penalties for poor alignment

35. Alignment (Cont.) • Example: A string “Hello” Takes 5 characters (without a terminating \0) • To word align next datum, add 3 “padding” characters to the string • The padding is not part of the string, it’s just unused memory